I was wondering if a matrix $A$ can be PD, but still have complex eigenvalues (with nonzero imaginary part).
I know for example that any symmetric/Hermitian matrix has real eigenvalues, but for example $A=\begin{pmatrix} 1 & 1\\0 &1\end{pmatrix}$ is neither, yet is positive definite.
From linear algebra I also know that the following statements are equaivalent:
1) $A$ is PD if $z^TAz>0 \;\forall z \neq 0$.
2) $A$ is PD, then all eigenvalues $\lambda_i>0$.
3) $A$ is PD, then $z^TAz>cz^Tz$ for some $c>0$.
I'm mostly concerned abou the second statement, so if it would possible for a PD matrix to have complex eigenvalues, how does this relate to $\lambda>0$? Are we only considering the real part?
At this point I am wondering if these statements are even true for the complex case, since in classes you usually see examples like $z^TAz = c_1z_1^2+c_2z_2^2$, (considering some $A\in\mathbb{R}^{2\times 2}$) where $c_1$ and $c_2$ are some positive constants. This however works only for real $z$, since if for example $z = [i\quad 0]^T$, this argument already breaks down.
Looking at for example Wikipedia seems not very helpful, since this only considers Hermitian/symmetric matrices, which I am specifically not interested in.
Summarising
I am interested in the possibilities of having PD matrices with complex eigenvalues. I am not particularly interested in a simple yes or no (even though this would be a great starting point), but more in what the consequences of such a situation are. How does this relate to the above statements? Do I oversee some principles ideas which even contradict this being possible?